Telescoping Series and Integration of Trigonometric Functions

When dealing with trigonometric integrals, particularly those involving products of sine and cosine functions, a compelling technique involves the use of telescoping series. This article explores this technique by analyzing (sin(nx)cdotcsc(x)) and dissecting it into simpler components. The goal is to derive a general formula for integrating (sin(nx)) in terms of sums of sine functions, leveraging the power of telescoping series and trigonometric identities.

Introduction to Trigonometric Identities

Trigonometric identities, such as the sine difference formula, are fundamental in simplifying and solving complex trigonometric expressions. The key identity we will use is:

$$sin A - sin B 2 cosleft(frac{A B}{2}right) cdot sinleft(frac{A-B}{2}right)$$

This identity allows us to break down expressions involving sine functions into simpler components, a crucial step in establishing a general integration formula.

Case I: Even (n)

Consider the case where (n) is an even integer. We begin by expressing (sin(nx)) using the sine difference formula repeatedly:

$$sin(nx) sin(nx - 2x) cdot frac{sin(2x)}{sin(2x)} left(sin((n-2)x) - sin((n-4)x)right) cdot sin(2x) / sin(2x)$$

This process can be iterated until we reach (sin(2x)) and (sin(0)):

$$sin(nx) 2cos((n-1)x) cdot sin(x) cdot 2cos((n-3)x) cdot sin(x) cdots 2cos(x) cdot sin(x)$$

Factoring out (sin(x)), we get:

$$sin(nx) 2sin(x) left(cos((n-1)x) cos((n-3)x) cdots cos(x)right)$$

Flipping the series backwards, we have:

$$sin(nx) 2sin(x) sum_{r1}^{n/2} cos(2r-1)x$$

For even (n 2m), the expression becomes:

$$I_{2m} 2 sum_{k1}^m frac{sin(2k-1)x}{2k-1}C_1$$

Case II: Odd (n)

Now, consider the case where (n) is an odd integer. Using a similar approach, we can express (sin(nx)) as follows:

$$sin(nx) left(sin(nx - 2x) - sin((n-4)x)right) cdot sin(2x) / sin(2x) cdots sin(3x) - sin(x)$$

After simplifying, we get:

$$sin(nx) 2sin(x) sum_{r1}^{(n-1)/2} cos(2r)x$$

For odd (n 2m 1), the expression becomes:

$$I_{2m 1} 2 sum_{k1}^m frac{sin(2k)x}{2k} I_1 sum_{k1}^m frac{sin(2k)x}{k}C_2 x sum_{k1}^m frac{sin(2k)x}{k}C_2$$

Generalizing the Cases

Combining both cases, we can generalize the expression for (sin(nx)):

$$sin(nx) sin(n mod 2)x cdot 2 sum_{r1}^{lfloor n/2 rfloor} cos(2r-1) (n mod 2)x$$

When we multiply by (csc(x)), we get:

$$sin(nx) cdot csc(x) y cdot 2 sum_{r1}^{lfloor n/2 rfloor} cos(2r-1)yx$$

where (y n mod 2).

Integration and Conclusion

Using the above expressions, we can now integrate (sin(nx) cdot csc(x)) with respect to (x). The result is:

(boxed{int sin(nx) cdot csc(x) dx x cdot frac{1}{2} sum_{r1}^{lfloor n/2 rfloor} frac{sin(2r-1)yx}{2r-1y}})

Note that (y 0) if (2 mid n) and (1) otherwise.

Example Integrals

Here are a few examples to illustrate the integration process:

(int sin(5x) csc(x) dx frac{1}{2} x (cos 2x cos 4x)) (int sin(7x) csc(x) dx frac{1}{2} x (cos 2x cos 4x cos 6x)) (int sin(6x) csc(x) dx x (cos x cos 3x cos 5x)) (int sin(8x) csc(x) dx x (2 cos x cos 3x cos 5x cos 7x))